The sum of the first five terms of the series | vedclass

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(A) The given series is $3 + 4\frac{1}{2} + 6\frac{3}{4} + \dots = 3 + \frac{9}{2} + \frac{27}{4} + \dots$This is a Geometric Progression $(G.P.)$ whe

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$39\frac{9}{16}$

Vedclass · Answer

(A) The given series is $3 + 4\frac{1}{2} + 6\frac{3}{4} + \dots = 3 + \frac{9}{2} + \frac{27}{4} + \dots$This is a Geometric Progression $(G.P.)$ where the first term $a = 3$ and the common ratio $r = \frac{9/2}{3} = \frac{3}{2}$.The sum of the first $n$ terms of a $G.P.$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.For $n = 5$ terms:$S_5 = \frac{3 \left[ (\frac{3}{2})^5 - 1 ight]}{\frac{3}{2} - 1} = \frac{3 \left[ \frac{243}{32} - 1 ight]}{\frac{1}{2}}$$S_5 = 6 \left[ \frac{243 - 32}{32} ight] = 6 \left[ \frac{211}{32} ight] = \frac{3 	imes 211}{16} = \frac{633}{16}$Converting to a mixed fraction: $\frac{633}{16} = 39\frac{9}{16}$.

The sum of the first five terms of the series $3 + 4\frac{1}{2} + 6\frac{3}{4} + \dots$ will be

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